Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 1, Article 25, 2005
SOME INEQUALITIES FOR THE SINE INTEGRAL
STAMATIS KOUMANDOS
D EPARTMENT OF M ATHEMATICS AND S TATISTICS
T HE U NIVERSITY OF C YPRUS
P.O. B OX 20537
1678 N ICOSIA , CYPRUS.
skoumand@ucy.ac.cy
Received 21 December, 2004; accepted 25 February, 2005
Communicated by H. Gauchman
A BSTRACT. We establish several sharp inequalities involving the function Si(x) =
Rx
0
sin t
t
dt.
Key words and phrases: Sine integral function, Positive trigonometric integrals, Sharp inequalities.
2000 Mathematics Subject Classification. 26D15, 26D05, 33B10.
1. I NTRODUCTION
Let
Z
x
sin t
dt ,
t
0
be the sine integral function which plays an important role in various topics of Fourier analysis
(cf. [2]). In this article we prove that the function Si(x) satisfies the inequalities given in the
theorem below.
Si(x) =
Theorem 1.1. For all x ≥ 0 and y ≥ 0, we have
(1.1)
0 ≤ Si(x) + Si(y) − Si(x + y) ≤ 2 Si(π) − Si(2 π) = 2.285722 . . . .
Both bounds are sharp. We also have
(1.2)
0 ≤ Si(x) + Si(y) ≤ x + y
and
(1.3)
Si(x)
x
≤ , for x ≥ y > 0 .
Si(y)
y
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
251-04
2
S TAMATIS KOUMANDOS
Note that inequality (1.1) contains the sub-additive property of the function Si(x) and may
be viewed as a two-dimensional analogue of the classical inequality
0 ≤ Si(x) ≤ Si(π) = 1.8519 . . . ,
for all x ≥ 0.
Inequalities (1.2) and (1.3) are also sharp because
Si(x) = x + O(x3 ), as x → 0.
A special case of (1.2) is the following
(1.4)
0<
Si(x)
< 1, for x > 0.
x
discrete analogue of (1.1), where the function Si(x) is replaced by Fejér’s sums Sn (x) =
PThe
n
sin kx
, has been obtained in [1].
k=1
k
2. L EMMAS
For the proof of inequalities (1.1) to (1.3) we need the following elementary lemmas.
Lemma 2.1. We suppose that the function f has a continuous derivative on [0, ∞) and that
f (0) = 0. If xf 0 (x) ≤ f (x) for all x in [0, ∞) then for 0 ≤ t ≤ s, we have t f (s x) ≤
s f (t x) ≤ t x s f 0 (0) for all x ∈ [0, ∞).
Proof. We fix x in [0, ∞) and define
g(t) :=
f (t x)
, for t > 0,
t
and g(0) = x f 0 (0). Differentiating with respect to t we obtain
t2 g 0 (t) = txf 0 (tx) − f (tx).
It follows from this that g is decreasing on [0, ∞) therefore for 0 ≤ t ≤ s, we get g(s) ≤
g(t) ≤ g(0), which completes the proof of Lemma 2.1.
Lemma 2.2. For all x > 0 we have
d
dx
(2.1)
1
Si(x) < 0.
x
Proof. It is clear that (2.1) is equivalent to
(2.2)
Si(x) − sin x > 0, x > 0.
The function Si(x) attains its absolute minimum on [π, ∞) at x = 2π and Si(2π) = 1.4181 . . ..
Thus we have to prove (2.2) only for 0 < x < π. The function Si(x) is strictly increasing on this
interval and since Si(π/2) = 1.37 . . ., it remains to show that (2.2) is valid for 0 < x < π/2.
Let h(x) := Si(x) − sin(x). This function is strictly increasing on [0, π/2) because the
inequality h0 (x) > 0 is equivalent to x cot x < 1 which is clearly true for this range of x and
therefore the proof of (2.2) is complete.
Notice that (2.1) implies (1.4).
J. Inequal. Pure and Appl. Math., 6(1) Art. 25, 2005
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S OME I NEQUALITIES FOR THE S INE I NTEGRAL
3
3. P ROOF OF T HEOREM 1.1
It follows from Lemma 2.2 that the function f (x) = Si(x) satisfies the conditions of Lemma
2.1. Obviously f 0 (0) = 1. Therefore, for 0 ≤ t ≤ s, we have
(3.1)
t Si(s z) ≤ s Si(t z) ≤ t s z, for all z ≥ 0.
For x > 0, y > 0, setting z = x + y, t =
x
,
x+y
s = 1 in this inequality we obtain
x
Si(x + y) ≤ Si(x) ≤ x,
x+y
and similarly for z = x + y, t =
y
,
x+y
s = 1 we have
y
Si(x + y) ≤ Si(y) ≤ y .
x+y
From these inequalities we conclude (1.2) and the first inequality of (1.1). Inequality (1.3)
follows easily from (3.1) setting z = 1, t = y, s = x.
In order to prove the second inequality in (1.1) we distinguish the following cases:
a) x + y ≥ π and
b) 0 < x + y < π.
In the case a) we recall that the function Si(x) attains its absolute maximum on [0, ∞) at
x = π while its absolute minimum on [π, ∞) is attained at x = 2π. Hence in this case we have
Si(x) + Si(y) − Si(x + y) ≤ 2 Si(π) − Si(2 π) = 2.285722 . . . .
In the case b) we consider the following subcases:
b1) 0 < x + y ≤ π/4,
b2) π/4 < x + y ≤ π/2,
b3) π/2 < x + y < 3 π/4 and
b4) 3 π/4 < x + y < π,
keeping in mind that the function Si(x) is strictly increasing on [0, π].
In the case b1) we have
π = 1.5179 . . . ,
Si(x) + Si(y) − Si(x + y) ≤ 2 Si
4
in the case b2) we have
π π Si(x) + Si(y) − Si(x + y) ≤ 2 Si
− Si
= 1.9825 . . . ,
2
4
in the case b3) we have
π 3π
Si(x) + Si(y) − Si(x + y) ≤ 2 Si
− Si
= 2.10873 . . .
4
2
and finally in the case b4) we have
Si(x) + Si(y) − Si(x + y) ≤ 2 Si(π) − Si
3π
4
= 1.96412 . . . .
The numerical values of the function Si(x) have been calculated using Maple 8.
The proof of Theorem 1.1 is now complete.
Remark 3.1. Alternately, one can prove the inequalities in (1.1) using standard techniques from
multivariate calculus. Indeed, let
K(x, y) := Si(x) + Si(y) − Si(x + y) .
J. Inequal. Pure and Appl. Math., 6(1) Art. 25, 2005
http://jipam.vu.edu.au/
4
S TAMATIS KOUMANDOS
We first observe that
K(0, 0) = 0, K(x, 0) = 0, K(0, y) = 0.
Next, we assume that x > 0, y > 0. The system of equations
∂
∂
K(x, y) = 0,
K(x, y) = 0 ,
∂x
∂y
has as solutions the lattice points
(x, y) = (µ π, ν π), µ, ν ∈ N ,
and this follows from the properties of the function sin x/x. Using the Hessian matrix test we
conclude that
1) When µ is even and ν is odd or µ is odd and ν is even, the points (µ π, ν π) are saddle
points.
2) When µ is odd and ν is odd the function K(x, y) has a local maximum at (µ π, ν π).
3) When µ is even and ν is even the Hessian matrix test gives no information about the
nature of the points (µ π, ν π).
We deal with the case 3) separately.
It is easy to see that
Z x
sin t sin(t + y)
K(x, y) =
−
dt ,
t
t+y
0
therefore, for m, n = 1, 2, 3 . . ., we have
Z
K(2mπ, 2nπ) =
0
2mπ
1
1
−
t t + 2nπ
sin t dt.
It follows from this that
Z
0 < K(2mπ, 2nπ) <
0
π
1
1
−
t t + 2nπ
sin t dt < Si(π) = 1.8519 . . .
Next in the case 2) we obtain for m, n = 0, 1, 2 . . .,
Z (2m+1)π 1
1
K (2m + 1)π, (2n + 1)π =
+
sin t dt
t t + (2n + 1)π
0
Z π
1
1
≤
+
sin t dt
t t + (2n + 1)π
0
Z π
1
1
≤
+
sin t dt
t t+π
0
= 2 Si(π) − Si(2 π) = 2.285722 . . .
This yields (1.1).
Remark 3.2. Using Lemma 2.1, one can prove more general inequalities involving the function
Si(x). Indeed, for the function f (x) = (Si(x))α xβ the condition x f 0 (x) ≤ f (x) is equivalent
to
(3.2)
(1 − β)Si(x) − α sin x > 0, x > 0 .
This inequality is valid precisely when α + β ≤ 1 and α ≥ 0. To see this, suppose first that
(3.2) holds. Dividing by Si(x) and letting x → 0 we obtain the first condition. From (3.2) when
α + β → 1 we get α ≥ 0, taking into account (2.2). Conversely, when α + β ≤ 1 and α ≥ 0,
J. Inequal. Pure and Appl. Math., 6(1) Art. 25, 2005
http://jipam.vu.edu.au/
S OME I NEQUALITIES FOR THE S INE I NTEGRAL
5
inequality (3.2) follows from (2.2). Thus we obtain analogous results to inequalities (1.2), (1.3)
and to the first inequality in (1.1) for the function f (x) = (Si(x))α xβ .
Remark 3.3. Several other sharp inequalities of the type considered in this paper may be obtained using an appropriate function f (x), which satisfies the conditions of Lemma 2.1.
R EFERENCES
[1] H. ALZER
submitted.
AND
S. KOUMANDOS, Sub- and superadditive properties of Fejér’s sine polynomial,
[2] A. ZYGMUND, Trigonometric Series, 3rd ed., Cambridge University Press, 2002.
J. Inequal. Pure and Appl. Math., 6(1) Art. 25, 2005
http://jipam.vu.edu.au/